MATLAB: What you get with: (0:n-1)*n + 1:n

additionbugcolonMATLABplussumunexpected

What you get with:

n = 4;
(0:n-1)*n + 1:n % 1 2 3 4

I was expecting:

(0:n-1)*n + (1:n) % 1 6 11 16

Best Answer

Two things going on here.

1) Addition is done before the colon operator. Compare this:

 >> 1 + 1:4
 ans =
      2     3     4

and this:

 >> 1 + (1:4)
 ans =
      2     3     4     5

That means that (0:n-1)*n + 1 is evaluated to a vector and that vector becomes the first input to another colon operator.

2) When a vector is given to the colon operator, the first element is used:

 >> (1:4):3
 ans =
      1     2     3

Related Solutions

MATLAB: Different result when adding variables

You are doing two different operation sequences, so why the surprise that you get two different results?

First Case

>> 1:3 + [0 0 5]
ans =
 1   2   3

because of the operator precedence (plus is higher than colon) is equivalent to

>> 1: (3 + [0 0 5])
ans =
 1   2   3

which is equivalent to

>> 1:[3,3,8]
ans =
 1   2   3

and thus is just

>> 1:3
ans =
 1   2   3

because the colon operator only uses the first element of the vector and ignore the rest: "If you specify nonscalar arrays, MATLAB interprets j:i:k as j(1):i(1):k(1)."

Second Case

 >> x = 1:3;
 >> y = [0 0 5];
 >> x+y
 ans =
   1   2   8

performs an element-wise addition on two vectors.

Summary

Beginners often forget about operator precedence and to read the documentation...

MATLAB: N(x)

CODE1)

a=[0 1];                % you tell MATLAB a is a vector
for r=3:100             % the for loop you understand, right?
    a(r)=(r-1)+(r-2);   % you're attributing to the rth term of your vector a values of (r - 1) + (r - 2)
                        % the 1st iteration will be a(3) = (3 - 1) + (3 - 2); a(4) = (4 - 1) + (4 - 2); so on...
end

disp(a)

CODE2)

a=[0 1];                %from this line, a(1) = 0 and a(2) = 1.
for r=3:100 
    a(r)=a(r-1)+a(r-2); % the difference between this and code 1) is that you're performing
                        % the summation calling the indexes in the vector a, rather than r values.
                        % so a(3) = a(2) + a(1) = 0 + 1 = 1
                        % In code 1, a(3) = 2 + 1 = 3
end
disp(a)

Code 3) is the same as 2) Walter mentioned.

Hope that helps...

Cheers

Best Answer

Related Solutions

MATLAB: Different result when adding variables

MATLAB: N(x)

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